United States presidential election in Maryland, 1992

United States presidential election in Maryland, 1992

November 3, 1992

Nominee Bill Clinton George H.W. Bush Ross Perot Party Democratic Republican Independent Home state Arkansas Texas Texas Running mate Al Gore Dan Quayle James Stockdale Electoral vote 10 0 0 Popular vote 988,571 707,094 281,414 Percentage 49.80% 35.62% 14.18%

County Results   Clinton—>70%   Clinton—60-70%   Clinton—50-60%   Clinton—40-50%   Bush—40-50%   Bush—50-60%   Bush—60-70%   Bush—>70%   Perot—40-50%

President before election George H. W. Bush Republican Elected President Bill Clinton Democratic

The 1992 United States presidential election in Maryland took place on November 3, 1992, as part of the 1992 United States presidential election. Voters chose ten representatives, or electors to the Electoral College, who voted for President and Vice President.

Maryland was won by Governor Bill Clinton (D-Arkansas) with 49.80% of the popular vote over incumbent President George H.W. Bush (R-Texas) with 35.62%. Businessman Ross Perot (I-Texas) finished in third with 14.18% of the popular vote.^[1] Clinton ultimately won the national vote, defeating incumbent President Bush.^[2]

Results

United States presidential election in Maryland, 1992[1]
Party		Candidate	Votes	Percentage	Electoral votes
	Democratic	Bill Clinton	988,571	49.80%	10
	Republican	George H.W. Bush	707,094	35.62%	0
	Independent	Ross Perot	281,414	14.18%	0
	Libertarian	Andre Marrou	4,715	0.24%	0
	New Alliance	Lenora Fulani	2,786	0.14%	0
	N/A	Write-ins	466	0.02%	0
Totals			1,985,046	100.0%	10

References

1. "1992 Presidential General Election Results - Maryland". U.S. Election Atlas. Retrieved 8 June 2012.
2. "1992 Presidential General Election Results". U.S. Election Atlas. Retrieved 8 June 2012.